Thermochemistry
Study energy changes in chemical reactions: calorimetry, Hess's law, standard enthalpies of formation, bond energies, and stoichiometric enthalpy.
Energy, Systems, and Enthalpy
Thermochemistry studies the energy changes that accompany chemical reactions. Key concepts:
- The system is the part of the universe being studied (usually the reactants and products). Everything else is the surroundings.
- Energy is conserved: it is never created or destroyed, only transferred between the system and its surroundings, most often as heat in the reactions studied here.
- Heat (q) flows from hot to cold. When q is negative, heat flows out of the system (exothermic); when positive, heat flows in (endothermic).
Enthalpy (H) is the heat content of a system at constant pressure. For reactions at constant pressure, the enthalpy change ΔH equals the heat exchanged: ΔH = qp. Exothermic reactions have ΔH < 0; endothermic reactions have ΔH > 0.
Go deeperWhy this matters: the pharmacy sells both signs of delta H
An instant cold pack is an endothermic process in a plastic sleeve: squeeze it, ammonium nitrate dissolves into the water pouch, and the mixture pulls heat from its surroundings (your sprained ankle) as it absorbs energy. A single-use hand warmer is the opposite sign: iron powder oxidizes in air, an exothermic reaction released slowly enough to stay comfortably warm for hours. Same shelf, opposite thermochemistry. Every reaction you will label exothermic or endothermic in this topic is a candidate product for one shelf or the other.
Heat Capacity, Specific Heat, and q = mcΔT
The relationship between heat transferred and temperature change is governed by specific heat capacity (c) — the amount of heat needed to raise 1 gram of a substance by 1 °C.
The equation: q = mcΔT
- q = heat transferred (J)
- m = mass (g)
- c = specific heat (J/g·°C). Water’s specific heat is 4.184 J/g·°C.
- ΔT = Tfinal − Tinitial (°C)
If q is positive, the substance absorbed heat (temperature rose). If q is negative, the substance released heat (temperature dropped).
Heat capacity (C) is the heat needed to raise the temperature of an entire object by 1 °C: q = CΔT. Specific heat is heat capacity per gram.
Go deeperWhy this matters: water's specific heat runs the planet's thermostat
Water’s specific heat, 4.184 J/g·°C, is extraordinarily high: several times that of rock or sand. Consequences surround you. Beach sand scorches your feet while the ocean beside it stays cool: same sun, different c. Coastal cities have milder climates than inland ones at the same latitude because the sea absorbs and releases heat sluggishly. Car engines are cooled with water, and your body, mostly water, rides out temperature swings that would crash a less water-based machine. When a substance resists temperature change, q = mcΔT says its c is large; water’s is among the largest of common substances.
Go deeperTry it: how much energy boils a kettle?
How much heat does it take to bring 250 g of water (one mug) from 20.0 °C to 100.0 °C? Work it, then check below.
Answer: q = mcΔT = 250 g × 4.184 J/g·°C × 80.0 °C = 83,680 J ≈ 84 kJ. Two habits worth building here: ΔT is final minus initial (80.0 °C, and it is the same size in kelvin, so no conversion is needed for a temperature difference), and the answer’s scale is worth a glance: tens of kilojoules to heat a mug of water is why electric kettles draw more power than almost anything else on your counter.
Calorimetry
Calorimetry measures heat changes by using an insulated container (calorimeter). The principle: heat lost by one substance equals heat gained by another (conservation of energy).
In a coffee-cup calorimeter (constant pressure): a reaction occurs in solution, and the temperature change of the solution is measured. Then qrxn = −qsolution = −mcΔT. Dividing by moles gives ΔH per mole. Example: if 50.0 g of water rises 6.0 °C during a reaction, qwater = (50.0)(4.184)(6.0) = 1255 J, so qrxn = −1255 J.
Two common reaction types in a coffee-cup calorimeter. A neutralization between a strong acid and a strong base (HCl + NaOH, HBr + NaOH, HClO4 + KOH, …) releases approximately the same amount of heat per mole of water formed, because the net ionic reaction is the same in every case: H+(aq) + OH−(aq) → H2O(l), with ΔH ≈ −56 kJ/mol. Identify the limiting reactant from the millimole counts (volume in mL × molarity), multiply by the per-mole heat to get the total heat released, and apply q = msolution·c·ΔT to find the resulting temperature rise. A dissolution measurement uses the same calorimeter but reports the heat of solution (also called heat of solvation): a positive value means the dissolution is endothermic and the solution cools, a negative value means exothermic and the solution warms. Combustion reactions are usually run in a bomb calorimeter, not a coffee cup, because the gases produced require a sealed vessel.
In a bomb calorimeter (constant volume): a substance is burned in a sealed steel container surrounded by water inside an insulated jacket. The heat released by the reaction raises the temperature of both the water and the calorimeter hardware. The full energy balance is qrxn = −(mwater·cwater + Ccal)·ΔT, where Ccal is the calorimeter’s heat capacity for the hardware alone (J/K, applied to the entire calorimeter as a single object — not per gram). Some textbook problems quote a single “effective Ccal” that already absorbs the mwater·cwater term; in that case the formula simplifies to qrxn = −Ccal·ΔT. Read each problem to see which convention is being used.
Heat of combustion from a bomb-calorimeter run. Burn a known mass of the compound, measure ΔT of the water + calorimeter, compute the total heat absorbed Q = (mwater·cwater + Ccal)·ΔT, then divide by the moles of compound burned to get heat of combustion in kJ/mol. Combustion is exothermic, so ΔHcomb carries a negative sign even though the calorimetric Q is positive.
Mixing two substances at different temperatures. When a hot substance and a cold substance reach thermal equilibrium inside a perfect calorimeter, the heat lost by the hot substance equals the heat gained by the cold substance: mhot·chot·(Thot − Teq) = mcold·ccold·(Teq − Tcold). When both substances are the same (e.g., hot water poured into cold water), the specific-heat term cancels and the equilibrium temperature is the mass-weighted average:
Teq = (mhot·Thot + mcold·Tcold) / (mhot + mcold)
For temperature differences, °C and K are interchangeable. For the mass-weighted average above, °C and K give the same equilibrium temperature (the constant 273.15 offset cancels because every term shifts by the same amount), so use whichever scale the problem provides.
Determining Ccal empirically. If a problem gives no calorimeter heat capacity, you can find one by mixing two known masses of water at known temperatures and using the gap between the predicted equal-mass midpoint temperature and the observed equilibrium. The full energy balance becomes qhot = qcold + qcal — three terms instead of two — and you solve for Ccal directly.
Go deeperWhy this matters: the Calorie label is calorimetry
The energy values on food labels trace back to exactly the instrument in this section: samples burned in a bomb calorimeter, with the temperature rise of the surrounding water revealing the energy content. One food Calorie (capital C) is one kilocalorie: 4184 J, the energy to raise a kilogram of water by one degree Celsius. A 250-Calorie candy bar therefore carries about one megajoule, enough, in principle, to heat nearly seven liters of ice-cold water to body temperature. Modern labels use standardized per-gram factors for fat, carbohydrate, and protein, but those factors were themselves established by calorimetry.
Enthalpy Diagrams
An enthalpy diagram plots the energy of reactants and products on a vertical axis:
- Exothermic reactions: products are lower in energy than reactants. The arrow points downward, and ΔH is negative. Energy is released to the surroundings.
- Endothermic reactions: products are higher in energy than reactants. The arrow points upward, and ΔH is positive. Energy is absorbed from the surroundings.
Go deeperCommon mistake: reading the negative sign as loss of importance
Students see ΔH = −890 kJ and hesitate: the reaction releases energy, so why is the number negative? Because the sign is written from the system’s ledger: the reacting chemicals lost that enthalpy to the surroundings. Exothermic means the system’s account went down (negative ΔH) while the surroundings warmed up; endothermic means the system banked energy (positive ΔH) at the surroundings’ expense.
Fix the perspective and every sign in the topic falls in line: the diagram’s downhill arrow, the “+ heat” written on the product side of an exothermic equation, and the thermostat logic of the cold pack and hand warmer.
Hess's Law
Hess’s law states that the enthalpy change for an overall reaction is the sum of the enthalpy changes for the individual steps, regardless of the pathway taken. This is because enthalpy is a state function — it depends only on the initial and final states.
To use Hess’s law: (1) Arrange given equations so that when added, reactants and products match the target equation. (2) If an equation is reversed, change the sign of ΔH. (3) If an equation is multiplied by a factor, multiply ΔH by that factor. (4) Add all ΔH values.
This method lets us calculate ΔH for reactions that are difficult to measure directly by combining reactions that are easy to measure.
Go deeperEnthalpy is an elevation, not a route
Hess’s law is the chemistry of a hiking fact: your net elevation gain from trailhead to summit is fixed by those two points, no matter which trail you take. Enthalpy is the elevation; the reaction pathway is the trail.
The power move this enables: measuring reactions that refuse to run cleanly. Nobody can burn diamond into graphite in a calorimeter and wait geological ages for the conversion, but both forms burn readily to CO2. Subtract the two combustion enthalpies and Hess’s law hands you the diamond-to-graphite value (about −2 kJ/mol) without the impossible experiment. Chemists compute unreachable enthalpies from reachable ones this way constantly.
Standard Enthalpies of Formation
The standard enthalpy of formation (ΔH°f) is the enthalpy change when exactly 1 mole of a compound is formed from its elements in their standard states (pure form at 25 °C and 1 atm). For example, the standard state of oxygen is O2(g), carbon is C(s, graphite), iron is Fe(s), and sulfur is S8(s, rhombic) — which is why a formation equation for H2SO4(l) takes the form H2(g) + ⅛S8(s) + 2 O2(g) → H2SO4(l) (fractional coefficients are required to deliver exactly one mole of the target compound). By convention, ΔH°f for any element already in its standard state is exactly zero.
This convention makes formation values powerful: the enthalpy of any reaction can be calculated from a table of ΔH°f values using:
ΔH°rxn = ∑ n·ΔH°f(products) − ∑ n·ΔH°f(reactants)
where n is the stoichiometric coefficient of each species in the balanced equation. This relationship follows directly from Hess’s law — every reaction can be imagined as decomposing reactants into elements, then recombining those elements into products.
Example: For CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l), ΔH° = [ΔH°f(CO2) + 2·ΔH°f(H2O)] − [ΔH°f(CH4) + 2·ΔH°f(O2)]. Because ΔH°f(O2) = 0, only three table values are needed.
Go deeperFormation enthalpies are sea level for chemistry
Elevations only mean something against an agreed zero, and geographers chose sea level. Thermochemists made the same move: elements in their standard states are declared zero, and every compound’s ΔH°f is its enthalpy “altitude” above or below that elemental sea level. Water sits at −285.8 kJ/mol: deep below sea level, which is another way of saying hydrogen and oxygen fall a long way down when they combine.
This is why the products-minus-reactants formula works: subtracting altitudes referenced to the same sea level gives the true climb between any two points, and the arbitrary zero cancels out of every calculation.
Bond Energies and Stoichiometric Enthalpy Calculations
Bond energy is the energy required to break one mole of a particular bond in the gas phase. It can be used to estimate ΔH for gas-phase reactions:
ΔH ≈ ∑(bonds broken) − ∑(bonds formed)
Breaking bonds requires energy (positive), forming bonds releases energy (negative). If more energy is released in forming new bonds than consumed in breaking old ones, the reaction is exothermic.
The reverse direction: extracting one bond energy. The same accounting can be run backward to find a single bond energy from formation data. Build a thermochemical cycle from free gas-phase atoms: the enthalpy to atomize the elements (break them into gaseous atoms), minus the compound’s ΔH°f, equals the total energy of the bonds holding the compound together. For HCl(g), half an H–H bond (½ × 436 kJ/mol) plus half a Cl–Cl bond (½ × 243 kJ/mol) minus ΔH°f(HCl, g) = −92.3 kJ/mol gives E(H–Cl) ≈ 432 kJ/mol. When the molecule contains several identical bonds — the two C=S bonds in CS₂, for example — divide the cycle total by the bond count.
Stoichiometric enthalpy calculations use ΔH as a conversion factor, just like mole ratios. If ΔH = −890 kJ for the combustion of 1 mol CH4, then burning 2.5 mol produces q = 2.5 × (−890) = −2225 kJ. You can also convert between grams, moles, and energy.
Go deeperCommon mistake: flipping the bond-energy formula
Two formulas in this topic point in opposite directions, and mixing them up flips every sign. With formation enthalpies, it is products minus reactants. With bond energies, it is broken minus formed: the bonds broken are in the reactants, so the subtraction runs reactants-first. Students who autopilot “products minus reactants” into a bond-energy problem get an answer that is exactly wrong in sign.
Anchor the logic physically instead of memorizing the order: breaking bonds always costs energy (positive), forming bonds always pays energy back (negative). If your exothermic combustion comes out positive, the formula got flipped, and the physical anchor catches it.
Thermochemistry Problem Selection and Common Mistakes
Choose the right approach by identifying what information is given:
- Temperature change measured? Use q = mcΔT (calorimetry).
- Standard enthalpies of formation available? Use ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants).
- Need to combine reactions? Use Hess’s law: reverse, multiply, and add reactions until the target equation is obtained.
- Bond energies given? Use ΔH ≈ Σ(bonds broken) − Σ(bonds formed). This is approximate and applies to gas-phase reactions.
- Stoichiometric conversion? ΔH applies to the equation as written. Scale it with the mole ratio.
Common mistakes: confusing the sign of ΔH (negative = exothermic, positive = endothermic), forgetting that ΔH°f for elements in their standard state is zero, using bond energies for condensed-phase reactions where they don’t apply well, reversing a reaction without changing the sign of ΔH, and using mass of solution instead of mass of solute (or vice versa) in calorimetry.
Key Equations
Learning Objectives
After studying this topic, you should be able to:
- Calculate heat transfer for a temperature change using specific heat or heat capacity
- Perform calorimetry calculations for coffee-cup and bomb calorimeters
- Apply Hess's law to calculate enthalpy changes for multi-step processes
- Calculate standard enthalpy of reaction from standard enthalpies of formation
- Perform stoichiometric calculations involving enthalpy (heat as a stoichiometric quantity)
- Estimate enthalpy of reaction from average bond energies
How-To Procedure
How to Calculate Enthalpy of Reaction Using Standard Enthalpies of Formation
- Write and balance the chemical equation.
- Look up the standard enthalpy of formation (ΔH°f) for each reactant and product from a reference table.
- Remember that ΔH°f for any element in its standard state is zero.
- Multiply each ΔH°f by its stoichiometric coefficient in the balanced equation.
- Apply the formula: ΔHrxn = Σ(n × ΔH°f products) − Σ(n × ΔH°f reactants).
- Check the sign: negative means exothermic, positive means endothermic.
Worked Example
Calorimetry: Finding ΔH of a Reaction
When 4.00 g of NaOH dissolves in 100.0 g of water in a coffee-cup calorimeter, the temperature rises from 22.0 °C to 28.4 °C. Calculate ΔH for dissolving NaOH (in kJ/mol). Assume c = 4.184 J/g·°C for the solution.
- Calculate q absorbed by solution: q = mc∆T = (104.0 g)(4.184 J/g·°C)(28.4 − 22.0) = (104.0)(4.184)(6.4) = 2784 J.
- q for the dissolving process is opposite in sign: qrxn = −2784 J = −2.784 kJ (exothermic — temperature rose).
- Calculate moles of NaOH: 4.00 g ÷ 40.00 g/mol = 0.100 mol.
- Calculate ΔH per mole: ΔH = −2.8 kJ ÷ 0.100 mol = −28 kJ/mol. The 6.4 °C temperature change limits the result to 2 significant figures.
ΔH = −28 kJ/mol (2 significant figures). The dissolution of NaOH is exothermic.
Test Your Understanding
A student mixes a hot metal block with cold water in a calorimeter and measures the final temperature. She calculates q(metal) and q(water) and finds that q(metal) + q(water) is not exactly zero. Does this mean energy was created or destroyed? What is the most likely explanation?
Self-Study Questions
What is the difference between an exothermic and an endothermic process?
Hint: Think about the sign of ΔH.
What is enthalpy and what does ΔH represent?
What is the relationship q = mcΔT and what does each variable mean?
What is calorimetry and how does a coffee-cup calorimeter work?
What is Hess’s law and why does it work?
What is the standard enthalpy of formation and what is its symbol?
Why is ΔH°f for any element in its standard state equal to zero?
How do you calculate ΔH°rxn from standard enthalpies of formation?
What is bond energy and how can it be used to estimate ΔH for a gas-phase reaction?
How do you use ΔH as a stoichiometric conversion factor?
When two amounts of the same liquid at different temperatures are mixed, how do you find the equilibrium temperature?
Hint: Mass-weighted average of the two starting temperatures.
Why is the heat of neutralization for any strong acid + strong base reaction approximately the same per mole of water formed?
How does a calorimeter's own heat capacity enter the energy balance, and how is it determined experimentally?
Content Sources
Concept sections draw on the sources listed below. OpenStax material is available under the CC BY 4.0 license.